let the time is T let translate the whole problem into a runner problem on the perimeter of a circle let consider unit circle at (0,0) . let call the point (1,0) A, (-1,0) B call the point (0,0) O, then we reformulate the problem in the following way the runners are starting from A runnin with different constant angular velocity going to the end point B in anticlockwise orientation, when they reach B they don't stop but go on same way in anticlock wise orientation to return to A. that mean there is some point X on the arc(which lies in the upper half plane) joinning AB and some point Y on the other arc (which lies in the lower half plane)joinning AB. and allthe runners are at one of the two point at time T. see those who are at X make angle $ \theta $(wrt to positive direction of x axis) then others make angle $ 2\pi - \theta $ now easily see at time 2T those who were at X now at a point making angle $ 2\theta $ and others are then at a point making angle $ 4\pi - 2\theta $ = $ 2\pi - 2\theta $ so they again meet by our formulation.